/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A potentiometric electrode for H... [FREE SOLUTION] | 91Ó°ÊÓ

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A potentiometric electrode for HCN uses a gas-permeable membrane, a buffered internal solution of \(0.01 \mathrm{M} \mathrm{KAg}(\mathrm{CN})_{2},\) and a \(\mathrm{Ag}_{2} \mathrm{~S}\) ISE electrode that is immersed in the internal solution. Consider the equilibrium reactions that take place within the internal solution and derive an equation that relates the electrode's potential to the concentration of \(\mathrm{HCN}\) in the sample.

Short Answer

Expert verified
The electrode potential is related to HCN concentration by: \[ E = E^0 - \frac{RT}{F} \ln \left(\frac{K_1 a_{\text{H}^+}^2}{K_2^2 a_{\text{HCN}}^2}\right) \]

Step by step solution

01

Write the relevant chemical equilibrium reactions

In the internal solution, the following equilibrium reactions are taking place: 1. Dissolution of potassium argentocyanide: \[ \text{KAg(CN)}_2 \rightleftharpoons \text{K}^+ + \text{Ag(CN)}_2^- \]2. Dissociation of silver-cyanide complex: \[ \text{Ag(CN)}_2^- \rightleftharpoons \text{Ag}^+ + 2 \text{CN}^- \]3. Hydrolysis of hydrogen cyanide in the sample: \[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]
02

Derive the Nernst equation for the Ag+ ion concentration

The potential of the silver ion-selective electrode (ISE) is given by the Nernst equation:\[ E = E^0 - \frac{RT}{nF} \ln a_{\text{Ag}^+} \]where \( E \) is the electrode potential, \( E^0 \) is the standard electrode potential, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of electrons exchanged (which is 1 for Ag), \( F \) is Faraday's constant, and \( a_{\text{Ag}^+} \) is the activity of silver ions.
03

Express silver ion concentration in terms of cyanide ions

The equilibrium involving \( \text{Ag(CN)}_2^- \rightleftharpoons \text{Ag}^+ + 2 \text{CN}^- \) can be used to relate \( a_{\text{Ag}^+} \) to \( a_{\text{CN}^-} \). The equilibrium constant \( K_1 \) for this reaction is:\[ K_1 = \frac{a_{\text{Ag}^+} a_{\text{CN}^-}^2}{a_{\text{Ag(CN)}_2^-}} \]Assuming the activity coefficient is approximately constant, you can solve for \( a_{\text{Ag}^+} \):\[ a_{\text{Ag}^+} = \frac{K_1}{a_{\text{CN}^-}^2} \]
04

Incorporate HCN concentration using its equilibrium

The dissociation of HCN gives us another relationship:\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]With an equilibrium constant \( K_2 = \frac{a_{\text{H}^+} a_{\text{CN}^-}}{a_{\text{HCN}}} \). Solve for \( a_{\text{CN}^-} \):\[ a_{\text{CN}^-} = \frac{K_2 a_{\text{HCN}}}{a_{\text{H}^+}} \]
05

Combine into a single expression for the electrode potential

Substitute \( a_{\text{CN}^-} \) from Step 4 into the expression for \( a_{\text{Ag}^+} \) in Step 3:\[ a_{\text{Ag}^+} = \frac{K_1}{(\frac{K_2 a_{\text{HCN}}}{a_{\text{H}^+}})^2} \]This simplifies to:\[ a_{\text{Ag}^+} = \frac{K_1 a_{\text{H}^+}^2}{K_2^2 a_{\text{HCN}}^2} \]Substitute this back into the Nernst equation:\[ E = E^0 - \frac{RT}{F} \ln \left(\frac{K_1 a_{\text{H}^+}^2}{K_2^2 a_{\text{HCN}}^2}\right) \]
06

Simplify the equation

Rearrange the terms and simplify, recognizing that some constants will combine:\[ E = E^0 - \frac{RT}{F} \left(\ln K_1 + 2 \ln a_{\text{H}^+} - 2 \ln K_2 - 2 \ln a_{\text{HCN}}\right) \]\[ E = E^0 - \frac{RT}{F} \ln K_1 + \frac{2RT}{F} \ln a_{\text{H}^+} - \frac{2RT}{F} \ln K_2 - \frac{2RT}{F} \ln a_{\text{HCN}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravimetric Analysis
Gravimetric analysis is a method used in chemistry to determine the quantity of an analyte based on the mass. It's a straightforward and highly accurate technique, which makes it widely employed in laboratories. To utilize this analysis, a chemical reaction is used to convert the analyte into a compound of known composition that can be isolated and weighed.
Consider the scenario where the analyst aims to determine the concentration of hydrogen cyanide (HCN). In the exercise, we see this principle in action, where the balance of masses and concentrations leads to understanding equilibrium states within the internal solution.
Gravimetric analysis can be broken down into key steps:
  • Selection of the appropriate reaction that forms a precipitate. This involves choosing a reaction where the precipitate can be easily filtered and weighed.
  • Performing the reaction and carefully isolating the solid precipitate.
  • Drying the precipitate completely and measuring its mass accurately.
  • Using the known chemical formula and stoichiometry to perform calculations that relate the mass of the precipitate to the quantity of the original analyte.
The accuracy of gravimetric analysis relies heavily on the purity of the precipitate, consistent drying methods, and precise weight measurement.
Nernst Equation
The Nernst Equation is a fundamental relationship in electrochemistry that explains how the potential of an ion-selective electrode changes with the concentration of the ions it detects. This concept is closely tied to the overall theme of electrode potentials, as showcased in the exercise.
In simple terms, the Nernst Equation for a half-cell reaction is expressed as:\[E = E^0 - \frac{RT}{nF} \ln (\text{activity})\]where:
  • \(E\) is the electrode potential.
  • \(E^0\) is the standard electrode potential.
  • \(R\) is the universal gas constant.
  • \(T\) is the absolute temperature in Kelvin.
  • \(n\) is the number of moles of electrons exchanged in the electrode reaction.
  • \(F\) is Faraday’s constant.
  • The "activity" represents the effective concentration of ionic species.
Comprehending the Nernst Equation helps in linking the chemical activities, which involve reaction dynamics, to the measurable quantity like potential.
In practical uses, the question often relates the equation to derive relations involving unknown concentrations, such as HCN in this example. By modifying the Nernst Equation with given conditions, one can deduce changes in electrode potentials.
Equilibrium Reactions
Equilibrium reactions are essential in understanding how chemical reactions behave under certain conditions. A chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction. As a result, the concentrations of reactants and products remain constant over time.
In the context of the exercise, multiple equilibrium reactions occur within the potentiometric system to derive electrode potential based on HCN concentration. These can be illustrated as follows:
  • The dissolution and dissociation reactions outline how chemical species break apart and form new interactions within the solution.
  • The hydrolysis of HCN is a key equilibrium, showcasing how it forms its ions and contributes to the overall potential equation.
At equilibrium, the relationship of concentrations of products and reactants is defined by an equilibrium constant (\(K\)). For example, in the HCN dissociation: \[\text{HCN} ightleftharpoons \text{H}^+ + \text{CN}^- \]the equilibrium constant \(K_2\) is given by: \[K_2 = \frac{a_{\text{H}^+} a_{\text{CN}^-}}{a_{\text{HCN}}} \]Understanding these principles emphasizes predicting how the system responds to changes.
Severity in shifts, such as concentration changes, produces notable effects, which is vital for advanced calculation and reactions planning.

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Most popular questions from this chapter

Mifflin and associates described a membrane electrode for the quantitative analysis of penicillin in which the enzyme penicillinase is immobilized in a polyacrylamide gel coated on the glass membrane of a \(\mathrm{pH}\) electrode. \({ }^{22}\) The following data were collected using a set of penicillin standards. \begin{tabular}{cc} [penicillin] (M) & potential (mV) \\ \hline \(1.0 \times 10^{-2}\) & 220 \\ \(2.0 \times 10^{-3}\) & 204 \\ \(1.0 \times 10^{-3}\) & 190 \\ \(2.0 \times 10^{-4}\) & 153 \\ \(1.0 \times 10^{-4}\) & 135 \\ \(1.0 \times 10^{-5}\) & 96 \\ \(1.0 \times 10^{-6}\) & 80 \end{tabular} (a) Over what range of concentrations is there a linear response? (b) What is the calibration curve's equation for this concentration range? (c) What is the concentration of penicillin in a sample that yields a potential of \(142 \mathrm{mV?}\)

The concentration of vanadium (V) in sea water is determined by adsorptive stripping voltammetry after forming a complex with catechol. \(^{3}\) The catechol-V(V) complex is deposited on a hanging mercury drop electrode at a potential of \(-0.1 \mathrm{~V}\) versus a \(\mathrm{Ag} / \mathrm{AgCl}\) reference electrode. A cathodic potential scan gives a stripping peak that is proportional to the concentration of \(\mathrm{V}(\mathrm{V})\). The following standard additions are used to analyze a sample of seawater. \begin{tabular}{cc} {\([\mathrm{V}(\mathrm{V})]_{\text {added }}(\mathrm{M})\)} & peak current \((\mathrm{n} \mathrm{A})\) \\ \hline \(2.0 \times 10^{-8}\) & 24 \\ \(4.0 \times 10^{-8}\) & 33 \\ \(8.0 \times 10^{-8}\) & 52 \\ \(1.2 \times 10^{-7}\) & 69 \\ \(1.8 \times 10^{-7}\) & 97 \\ \(2.8 \times 10^{-7}\) & 140 \end{tabular} Determine the molar concentration of \(\mathrm{V}(\mathrm{V})\) in the sample of sea water, assuming that the standard additions result in a negligible change in the sample's volume.

The concentration of \(\mathrm{H}_{2} \mathrm{~S}\) in the drainage from an abandoned mine is determined by a coulometric titration using KI as a mediator and \(I_{3}^{-}\) as the titrant. $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{I}_{5}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)=2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+3 \mathrm{I}^{-}(a q)+\mathrm{S}_{(s)} $$ A \(50.00-\mathrm{mL}\) sample of water is placed in a coulometric cell, along with an excess of KI and a small amount of starch as an indicator. Electrolysis is carried out at a constant current of \(84.6 \mathrm{~mA}\), requiring \(386 \mathrm{~s}\) to reach the starch end point. Report the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) in the sample in \(\mu \mathrm{g} / \mathrm{mL}\)

The polarographic half-wave potentials (versus the SCE) for \(\mathrm{Pb}^{2+}\) and for \(\mathrm{Tl}^{+}\) in \(1 \mathrm{M} \mathrm{HCl}\) are, respectively, \(-0.44 \mathrm{~V}\) and \(-0.45 \mathrm{~V}\). In an electrolyte of \(1 \mathrm{M} \mathrm{NaOH}\), however, the half-wave potentials are \(-0.76 \mathrm{~V}\) for \(\mathrm{Pb}^{2+}\) and \(-0.48 \mathrm{~V}\) for \(\mathrm{Tl}^{+}\). Why does the change in electrolyte have such a significant effect on the half-wave potential for \(\mathrm{Pb}^{2+}\), but not on the half- wave potential for \(\left.\mathrm{T}\right|^{+}\)

What reaction prevents us from using \(Z n\) as an electrode of the first kind in an acidic solution? Which other metals do you expect to behave in the same manner as Zn when immersed in an acidic solution?

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